Question

How many ions are present in 30.0mL of 0.600M Na2CO3 solution?

Answer 1

The correct answer to this question is this one:

Molarity = moles / volume in dm3 or litre

1st of all convert volume 30 mL into Litre or dm3 by dividing it with 1000. then it becomes 30/1000 = 0.03

Now  find  the  moles  by arranging the above equation, i.e
moles = molarity * volume
moles = 0.6 * 0.03
moles = 0.018

now  we have got the moles of Na2CO3

1 mole of Na2CO3 = 3 moles of total ions. (2NA+ and CO-2)
0.018 moles of Na2CO3 = 3*0.018
0.018 moles of Na2CO3 = 0.054 moles of ions

now apply  the  formula, i.e number of atoms or ions = moles * Na (avogadros number.)

number of ions = moles of ions * Na (avogadro's number)
number of ions = 0.054 * 6.022 * 10^{3}
number of ions =  3.025 * 10^{22}

Answer 2

$\boxed{{\text{3}}{\text{.25}}\times{\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ ions}}}$  are present in 30 mL of 0.600 M ${\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}$  solution.

Further Explanation:

The proportion of substance in the mixture is called concentration. The most commonly used concentration terms are as follows:

1. Molarity (M)

2. Molality (m)

3. Mole fraction (X)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M and its unit is mol/L.

The formula to calculate the molarity of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$] solution is as follows:

${\text{Molarity of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ solution}}=\frac{{{\text{Moles}}\;{\text{of}}\;{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}}{{{\text{Volume }}\left({\text{L}}\right){\text{ of}}\;{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ solution}}}}$       …… (1)

Rearrange equation (1) to calculate the moles of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$  solution.

$\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}=\left({{\text{Molarity of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ solution}}}\right)\\\left({{\text{Volume of}}\;{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ solution}}}\right)\\\end{aligned}$                                          …… (2)

The molarity of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$  solution is 0.600 M.

The volume of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$  solution is 30 mL.

Substitute these values in equation (2).

$\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}&=\left({{\text{0}}{\text{.600 M}}}\right)\left({{\text{30 mL}}}\right)\left({\frac{{{{10}^{ - 3}}{\text{ L}}}}{{{\text{1 mL}}}}}\right)\\&=0.01{\text{8 mol}}\\\end{aligned}$

The dissociation of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$  occurs as follows:

${\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}\rightleftharpoons 2{\text{N}}{{\text{a}}^ + } + {\text{CO}}_3^{2 - }$

This indicates that one mole of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$  dissociates to form two moles of ${\text{N}}{{\text{a}}^ + }$  ions and one mole of ${\text{CO}}_3^{2 - }$   ion. So total of three moles of ions are produced by one mole of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ .

The total number of moles of ions produced during the reaction is calculated as follows:

$\begin{aligned}{\text{Moles of ions}}&=\left({{\text{0}}{\text{.018 mol}}}\right)\left({\text{3}} \right)\\&=0.0{\text{54 mol}}\\\end{aligned}$

The formula to calculate the number of ions is as follows:

${\text{Number of ions}}=\left({{\text{Moles of ions}}}\right)\left({{\text{Avogadro's Number}}}\right)$                         …… (3)

The number of moles of ions is 0.054 mol.

The value of Avogadro’s number is ${\text{6}}{\text{.022}}\times{\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{ions}}$ .

Substitute these values in equation (3).

$\begin{aligned}{\text{Number of ions}}{\mathbf{ &= }}\left({0.054{\text{ mol}}}\right)\left( {\frac{{{\text{6}}{\text{.022}}\times{\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ ions}}}}{{{\text{1 mol}}}}} \right)\\&={\text{3}}{\text{.25188}}\times{\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ ions}}\\&\approx {\text{3}}{\text{.25}}\times{\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ ions}}\\\end{aligned}$

Learn more:

1. The mass of ethylene glycol: brainly.com/question/4053884

2. Calculate the moles of ions in HCl solution: brainly.com/question/5950133

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: molarity, ions, moles, Na2CO3, Na+, CO32-, 2Na+, 0.018 mol, 0.054 mol, Avogadro’s number, 3.25*10^22 ions, volume of Na2CO3, moles of Na2CO3.